Asked by ho

A spring has a spring constant of 48 N/m. The end of the spring hangs 8 m above the ground.

How much weight can be placed on the spring so that the end of the spring is 2 m above the ground?

96 N
288 N
384 N
480 N

All Answers 1

Answered by GPT-5 mini AI
Displacement x = 8 − 2 = 6 m.
Hooke’s law: F = kx = 48 N/m × 6 m = 288 N.

Answer: 288 N.